WORK, ENERGY AND POWER

WORK, ENERGY AND POWER 4.1 Introduction Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force. 4. Work Done by a Constant Force Let a constant force F be applied on the body such that it makes an angle 6 with the horizontal and body is displaced through a distance s Then work done by the force in displacing the body through a distance a is given by W = (F cos q)s = Fs cos q W = (F cos q)s = Fs cos q W = F. s 4.3 Nature of Work Done Positive work Positive work means that force (or its component) is parallel to displacement 0 o q < 90 o F S Direction of motion The positive work sigmnes that the external force favours the motion of the body. Negative work Negative work means that force (or its component) is opposite to displacement i.e. 90 o < q 180 o F Direction of motion The negative work sigmnes that the external force opposes the motion of the body. S

4.4 Work Done by a Variable Force When the magnitude and direction of a force varies with position, the work done by such a force for an ininitesimal displacement is given by dw = F. ds The total work done in going from A to B is W = B A F B. ds = A (F cos )ds Area under force displacement curve with proper algebraic sign represents work done by the force. 4.5 Work Depends on Frame of Reference. With change of frame of reference (inertial) force does not change while displacement may change. So the work done by a force will be different in different frames. Examples: If a person is pushing a box inside a moving train, the work done in the frame of train will F.s while in the frame of earth will be F.(s + s 0) where s 0 is the displacement of the train relative to the ground. 4.6 Energy The energy of a body is deined as its capacity for doing work. (1) It is a scalar quantity. () Dimension: [ML T ] it is same as that of work or torque. (3) Units: Joule [S.I.], erg [C.G.S.] Practical units: electron volt (ev), Kilowatt hour (KWh), Calories (Cal) Relation between different units: 1 Joule = 10 7 erg 1 ev= 1.6 10 19 Joule 1 KWh=3.6 10 6 Joule 1 Calorie =4.18 Joule (4) Mass energy equivalence: The relation between the mass of a particle m and its equivalent energy is given as E = mc where c = velocity of light in vacuum.

4.7 Kinetic Energy The energy possessed by a body by virtue of its motion is called kinetic energy. Let m = mass of the body, v= velocity of the body W = mv (1) Kinetic energy depends on frame of reference: The kinetic energy of a person of mass m, sitting in a train moving with speed v, is zero in the frame of train but 1 mv in the frame of the earth. () Work-energy theorem: It states that work done by a force acting on a body is equal to the change produced in the kinetic energy of the body. This theorem is valid for a system in presence of all types of forces (external or internal, conservative or non-conservative). (3) Relation of kinetic energy with linear momentum: As we know Momentum P = E v = me (4) Various graphs of kinetic energy 4.8 Potential Energy E E E α P E α v m = constant m = constant v n E 1 E α m P = constant m E p α E m = constant P

4.9 Potential Energy Potential energy is deined only for conservative forces. In the space occupied by conservative forces every point is associated with certain energy which is called the energy of position or potential energy. Potential energy generally are of three types: Elastic potential energy, Electric potential energy and Gravitational potential energy etc. (1) Change in potential energy: Change in potential energy between any two points is deined in the terms of the work done by the force in displacing the particle between these two points without any change in kinetic energy. r U U 1 = F.d r = W...... (i) r1 () Potential energy curve- A graph plotted between the potential energy of a particle and its displacement from the centre of force is called potential energy curve. Negative gradient of the potential energy gives force. du = F (5) Types of equilibrium: If net force acting on a particle is zero, it is said to be in equilibrium. For equilibrium = du types: = 0, but the equilibrium of particle can be of three Stable Unstable Neutral When a particle is displaced slightly from a position, then a force acting on it brings it back to the initial position, it is said to be in stable equilibrium position. When a particle is displaced slightly from a position, then a force acting on it tries to displace the particle further away from the equilibrium position, it is said to be in unstable equilibrium. When a particle is slightly displaced from a position then it does not experience any force acting on it and continues to be in equilibrium in the displaced position, it is said to be in neutral equilibrium. Potential energy is minimum. Potential energy is maximum. Potential energy is constant. F = du = 0 F = du = 0 F = du = 0

Stable Unstable Neutral du = positive i.e. rate of change of du is positive. Example: A marble placed at the bottom of a hemispherical bowl. du = negative i.e. rate of change of du is negative Example: A marble balanced on top of a hemispherical bowl. du = 0 i.e. rate of change of du is zero. Example: A marble placed on horizontal table. 4.10 Elastic Potential Energy. (1) Restoring force and spring constant- When a spring is stretched or compressed from its normal position (x = 0) by a small distance x, a restoring force is produced in the spring to bring it to the normal position. According to Hooke s law this restoring force is proportional to the displacement x and its direction is always opposite to the displacement. i.e. F α x or F = k x...... (i) where k is called spring constant. K A Energy E O B x = a x = 0 x = + a Position u () Expression for elastic potential energy- :. Elastic potential energy U = 1 kx = 1 Fx = F k

Note: If spring is stretched from initial position x 1 to inal position x then work done = Increment in elastic potential energy 1 = k ( x x1 ) (3) Energy graph for a spring: It mean kinetic energy changes parabolically w.r.t. position but total energy remain always constant irrespective to position of the mass. 4.11 Law of Conservation of Energy. (1) Law of conservation of energy: For an isolated system or body in presence of conservative forces the sum of kinetic and potential energies at any point remains constant throughout the motion. It does not depends upon time. This is known as the law of conservation of mechanical energy. () Law of conservation of total energy: If the forces are conservative and non-conservative both, it is not the mechanical energy alone which is conserved, but it is the total energy, may be heat, light, sound or mechanical etc., which is conserved. 4.15 Power Power of a body is deined as the rate at which the body can do the work. W Average power (P av. ) = t = W. Instantaneous power (P t inst. ) = dw Fds. = dt dt [As dw = Fds. ] P inst = Fv.. i.e. power is equal to the scalar product of force with velocity. (1) Dimension: [P] = [MLT 3 ] () Units: Watt or Joule/sec [S.I.] Practical units: Kilowatt (kw), Mega watt (MW) and Horse power (hp) Relations between different units: 1 watt = 1 Joule/sec = 10 7 erg /sec 1hp = 746 Watt

(3) The slope of work time curve gives the instantaneous power. As P = dw/dt = tanq (4) Area under power time curve gives the work done as P = dw dt :. W = P dt 4.1 Collision :. W = Area under P t curve Collision is an isolated event in which a strong force acts between two or more bodies for a short time as a result of which the energy and momentum of the interacting particle change. In collision particles may or may not come in real touch (3) Types of collision: (i) On the basis of conservation of kinetic energy. Perfectly elastic collision Inelastic collision Perfectly inelastic collision If in a collision, kinetic energy after collision is equal to kinetic energy before collision, the collision is said to be perfectly elastic. If in a collision kinetic energy after collision is not equal to kinetic energy before collision, the collision is said to inelastic. Coeficient of restitution e = 1 Coeficient of restitution 0 < e < 1 (KE) inal = (KE) initial Here kinetic energy appears in other forms. In some cases (KE) inal < (KE) initial such as when initial KE is converted into internal energy of the product (as heat, elastic or excitation) while in other cases (KE) final > (KE) initial such as when internal energy stored in the colliding particles is released. If in a collision two bodies stick together or move with same velocity after the collision, the collision is said to be perfectly inelastic. Coeficient of restitution e = 0 The term perfectly inelastic does not necessarily mean that all the initial kinetic energy is lost, it implies that the loss in kinetic energy is as large as it can be. (Consistent with momentum conservation).

Perfectly elastic collision Inelastic collision Perfectly inelastic collision Examples: (1) Collision between atomic particles () Bouncing of ball with same velocity after the collision with earth. Examples: (1) Collision between two billiard balls. () Collision between two automobile on a road. In fact all majority of collision belong to this category. 4.13 Perfectly Elastic Head on Collision. Example: Collision between a bullet and a block of wood into which it is ired. When the bullet remains embeded in the block. Let two bodies of masses m 1 and m moving initial velocities u 1 and u in the same direction they collide such that after collision their inal velocities are v 1 and v respectively. u 1 u v 1 v m 1 m m 1 m Before collision After collision According to law of conservation of momentum and conservation of kinetic energy. Note: The ratio of relative velocity of separation and relative velocity of approach v v1 e = or v v1 = e(u is deined as coeficient of restitution. u1 u 1 u ) For perfectly elastic collision e = 1 :. v v1 = u 1 u (As shown in eq. (vi) For perfectly inelastic collision e = 0 :. v v1 = 0 or v v1 It means that two body stick together and move with same velocity. For inelastic collision 0 < e < 1 :. v v1 = (u 1 u ) In short we can say that e is the degree of elasticity of collision and it is dimension less quantity. v v m m mu m + m m + m 1 1=+ u1 1 1 m m mu m + m m + m 1 1 1 =+ u1 1 1........ (vii)........ (viii)

When two bodies of equal masses undergo head on elastic collision, their velocities get interchanged. () Kinetic energy transfer during head on elastic collision. Fractional decrease in kinetic energy K 4mm 1 = K m m + 4mm ( ) 1 1....... (iv) Note: Greater the difference in masses less will be transfer of kinetic energy and vice versa Transfer of kinetic energy in head on elastic collision (when target is at rest) is maximum when the masses of particles are equal..14 Motion in Vertical Circle. This is an example of non-uniform circular motion. In this motion body is under the influence of gravity of earth. (1) Velocity at any point on vertical loop- If u is the initial velocity imparted to body at lowest point then. Velocity of body at height h is given by = = ( ) v u gh u gl 1 cosq where 1 in the length of the string D l h C O q A B v P u () Tension at any point on vertical loop: Tension at general point P, T = m g cos q + mv l (3) Various conditions for vertical motion: v velocity at lowest point Condition ua > 5gl Tension in the string will not be zero at any of the point and body will continue the circular motion.

ua = 5 gl, Tension at highest point C will be zero and body will just complete the circle. gl < u < 5 gl, A Particle will not follow circular motion. Tension in string become zero somewhere between points B and C whereas velocity remain positive. Particle leaves circular path and follow parabolic trajectory. ua ua = < gl gl Both velocity and tension in the string becomes zero between A and B and particle will oscillate along semi-circular path. velocity of particle becomes zero between A and B but tension will not be zero and the particle will oscillate about the point A. (6) Various quantities for a critical condition in a vertical loop at different positions: Quantity Point A Point B Point C Point D Point P Linear velocity (v) 5gl 3gl gl 3gl gl ( 3 + cosq ) Angular velocity (ω) 5g l 3g l g l 3g l g l ( 3 + cosq ) Tension in String (T) 6 mg 3 mg 0 3 mg 3 mg(1 + cos q) Kinetic Energy (KE) 5 mgl 3 mgl 1 mgl 3 mgl mu 1 5mg = 0 Potential Energy (PE) 0 mgl mgl mgl Mgl(1 cos q) Total Energy (TE) 5 mgl 5 mgl 5 mgl 5 mgl 5 mgl VERY SHORT ANSWER QUESTION (MARK 1) 1. Deine the conservative and non conservative forces. Give examples of each.. A light body and a heavy body have same linear momentum. Which one has greater K.E? (Ans. : lighter body has more K.E.) 3. The momentum of the body is doubled what % does its K.E change? [Ans : (300%)]

4. A body is moving along a circular path. How much work is done by the centripetal force? 5. Which spring has greater value of spring constant - a hard spring or a delicate spring? 6. Two bodies stick together after collision. What type of collision is in b/w these two bodies? 7. State the two conditions under which a force does no work? 8. How will the momentum of a body changes if its K.E is doubled? 9. K.E of a body is increased by 300 %. Find the % increase in its momentum? (100%) 10. A light and a heavy body have same K.E. which of the two have more momentum and why? (heavier body) 11. Mountain roads rarely go straight up the slop, but wind up gradually. Why? 1. A truck and a car moving with the same K.E on a straight road. Their engines are simultaneously switched off which one will stop at a lesser distance? 13. What happens to the P.E of a bubble when it rises in water? (decrease) 14. A body is moving at constant speed over a friction surface. What is the work done by the weight of the body? (W = 0) 15. Is it necessary that work done in the motion of a body over a closed loop is zero for every force in nature? Why? 16. Deine spring constant of a spring? 17. What happens when a sphere collides head on elastically with a sphere of same mass initially at rest? SHORT ANSWER QUESTIONS ( MARKS) 18. A elastic spring is compressed by an amount x. Show that its PE. is 1/ kx Where k is the spring constant?

19. Derive an expression for it K.E of a body of mass m moving with a velocity v by calculus method. 0. Show that the total mechanical energy of a body falling freely under gravity is conserved. Show the variation of K.E., RE. and Total Energy with height from earth surface. 1. How high must a body be lifted to gain an amount of RE equal to the K.E it has when moving at speed 0 ms 1. (The value of acceleration due to gravity at a place is 9.8 ms ). (0. m). A car of mass 000 kg is lifted up a distance of 30 m by a crane in 1 min. A second crane does the same job in min. Do the cranes consume the same or different amounts of fuel? What is the power supplied by each crane? Neolect vower dissipation against friction. 3. Prove that bodies of identical masses exchange their velocities after head-on elastic collision. 4. A bob is pulled sideway so that string becomes parallel to horizontal and released. Length of the pendulum is m. If due to air resistance loss of energy is 10% what is the speed with which the bob arrived at the lowest point. 5. Two springs A and B are identical except that A is harder than B(K A > K B ) if these are stretched by the equal force. In which spring will more work be done? 6. Answer the following : (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere or both? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why? 7. Find the work done if a particle moves from position r 1 = (3î+ĵ 6ˆk) to a position r = (14î+13ĵ 9ˆk) under the effect of force. F = (4î+ĵ+3ˆk)N.

8. Deine elastic and inelastic collision. A lighter body collides with a much more massive body at rest. Prove that the direction of lighter body is reserved and massive body remains at rest. 9. A car of mass 1000 kg accelerates uniformly from rest to a velocity of 54 kh h 1 in 5 seconds. Calculate (i) its acceleration (ii) its gain in K.E. (iii) average power of the engine during this period, neglect friction. [Ans. (i) 3 ms (ii) 1.15 10 5 J (iii) 500 W] 30. A ball at rest is dropped from a height of 1 m. It loses 5% of its kinetic energy in striking the ground, ind the height to which it bounces. How do you account for the loss in kinetic energy? 31. 0 J work is required to stretch a spring through 0.1 m. Find the force constant of the spring. If the spring is further stretched through 0.1m. Calculate work done. (4000 Nm 1, 60 J) 3. A body of mass M at rest is struck by a moving body of mass m. Prove that fraction of the initial K.E. of the mass m transferred to the struck body is 4 m M/(m + M) in an elastic collision. 33. A pump on the ground floor of a building can pump up water to ill a tank of volume 30m 3 in 15 min. If the tank is 40 m above the ground, how much electric power is consumed by the pump. The eficiency of the pump is 30%. (43.567 KW) 34. How much work is done by a collie walking on a horizontal platform with a load on his head? 35. Spring A and B are identical except that A is stiffer than B, i.e., force constant k A > k B. In which spring is more work expended if they are stretched by the same amount? 36. State and prove Work Energy Theorem. 37. Show that in an elastic one dimensional collision the relative velocity of approach before collision is equal to the relative velocity of separation after collision. 38. Show that in a head on collision between two balls of equal masses moving along a straight line the balls exchange their velocities.

39. Which of the two kilowatt hour or electron volt is a bigger unit of energy and by what factor? 40. A spring of force constant K is cut into two equal pieces. Calculate force constant of each part. 41. It is necessary that work done in the motion of a body over a closed loop is zero for every force in nature? Why? 4. A spring is irst stretched by x by applying a force F. Now the extension of the spring is increases to 3x. What will be the new force required to keep the spring in this condition? Calculate the work done in increasing the extension. LONG ANSWER QUESTIONS (5 MARKS) 43. Show that at any instant of time during the motion total mechanical energy of a freely falling body remains constant. Show graphically the variation of K.E. and P.E. during the motion. 44. Deine spring constant, Write the characteristics of the force during the elongation of a spring. Derive the relation for the PE stored when it is elongated by X. Draw the graphs to show the variation of P.E. and force with elongation. 45. How does a perfectly inelastic collision differ from perfectly elastic collision? Two particles of mass m 1 and m having velocities U 1 and U respectively make a head on collision. Derive the relation for their inal velocities. Discuss the following special cases. (i) m 1 = m (ii) m 1 >> m and U = 0 (iii) m 1 << m and u 1 = 0 NUMERICALS 46. A body is moving along z-axis of a coordinate system under the effect of a constant force F = (î+3ĵ ˆk)N. Find the work done by the force in moving the body a distance of m along z-axis.

47. In lifting a 10 kg weight to a height of m, 30 J energy is spent. Calculate the acceleration with which it was raised? 48. A bullet of mass 0.0 kg is moving with a speed of 10ms 1. It can penetrate 10 cm of a wooden block, and comes to rest. If the thickness of the target would be 6 cm only ind the KE of the bullet when it comes out. (Ans : 0.4 J) 49. A man pulls a lawn roller with a force of 0 kg F. If he applies the force at an angle of 60 with the ground. Calculate the power developed if he takes 1 min in doing so. 50. A ball bounces to 80% of its original height. Calculate the mechanical energy lost in each bounce. 51. A pendulum bob of mass 0.1 kg is suspended by a string of 1 m long. The bob is displaced so that the string becomes horizontal and released. Find its kinetic energy when the string makes an angle of (i) 0, (ii) 30 with the vertical. 5. A truck of mass 1000 kg accelerates uniformly from rest to a velocity of 15ms 1 in 5 seconds. Calculate (i) its acceleration (ii) its gain in K.E. (iii) average power of the engine during this period, neglect friction. [Ans. (i) 3 ms (ii) 1.15 x 10 5 J (iii) 500 W] 53. An elevator which can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of ms 1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power. 54. The simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 kmh 1 on a smooth road and colliding with a horizontally mounted spring of spring constant 6.5 10 3 Nm 1. What is the maximum compression of the spring. 55. A ball at rest is dropped from a height of 1 m. It loses 5% of its kinetic energy in striking the ground, ind the height to which it bounces. How do you account for the loss in kinetic energy?

56. A body of mass 0.3 kg is taken up an inclined plane to length 10 m and height 5 m and then allowed to slide down to the bottom again. The coeficient of friction between the body and the plane is 0.15. What is the (i) work done by the gravitational force over the round trip. (ii) work done by the applied force over the upward journey. (iii) work done by frictional force over the round trip. (iv) kinetic energy of the body at the end of the trip. How is the answer to (iv) related to the irst three answer? 57. Two identical 5 kg blocks are moving with same speed of ms 1 towards each other along a frictionless horizontal surface. The two blocks collide, stick together and come to rest. Consider the two blocks as a system. Calculate work done by (i) external forces and (i) Internal forces.